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\(\hept{\begin{cases}2xy-8x-2y+4=0\\2x^2-4x+4=2y^2-16y+40\end{cases}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}xy-4x-y+2=0\\x^2-2x+2=y^2-8y+20\end{cases}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-1\right)\left(y-4\right)=2\\\left(x-1\right)^2-\left(y-4\right)^2=3\end{cases}}\)
Đặt \(x-1=a;\)\(y-4=b\)ta có hpt:
\(\hept{\begin{cases}ab=2\\a^2-b^2=3\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}a=\frac{2}{b}\\\frac{4}{b^2}-b^2=3\end{cases}}\)
P/S: mk lm đc vậy thôi, bn tham khảo nhé!
mk mới lớp 8 nên cx ko biết trình bày đúng hay sai
giải ra thì đc kết quả như của bn Nguyễn Xuân Anh
Giải PT: \(x^2+3y^2+2xy-8x-16y+23=0\)
\(\Leftrightarrow x^2+y^2+16+2xy-8x-8y+2y^2-8y+7=0\)
\(\Leftrightarrow\left(x+y-4\right)^2+2\left(y^2-4y+4\right)-1=0\)
\(\Leftrightarrow\left(x+y-4\right)^2+2\left(y-2\right)^2-1=0\)
\(\Rightarrow\left(x+y-4\right)^2=-2\left(y-2\right)^2+1\le1\)
Dấu "=" xảy ra khi : \(-2\left(y-2\right)^2=0\Rightarrow y=2\)
\(\Rightarrow\)\(\text{│}x+y-4\text{│}\le1\)
\(\Rightarrow-1\le x+y-4\le1\)
\(\Rightarrow3\le x+y\le5\)
Vậy Bmin=3 khi y=2;x=1
Bmax=5 khi y=2;x=3
1) x2 + xy-8x-8y
= x.(x+y) -8.(x+y)
= (x+y).(x-8)
2) 9x2 -6x + 1 -36y2
=(3x-1)2 -(6y)2
=(3x-1-6y)(3x-1+6y)
3)a2 - b2 -12a+12b
= (a-b)(a+b)-12.(a-b)
=(a-b).(a+b-12)
4)x2 -2xy+8x-16y
=x.(x-2y)+8.(x-2y)
=(x-2y).(x+8)
5) x2 -9y2 +5x+15y
=(x-3y)(x+3y)+5.(x+3y)
= (x+3y).(x-3y+5)
\(1,x^2+xy-8x-8y=x\left(x+y\right)-8\left(x+y\right)=\left(x-8\right)\left(x+y\right)\)\(3,a^2-b^2-12a+12b=\left(a-b\right)\left(a+b\right)-12\left(a-b\right)=\left(a-b\right)\left(a+b-12\right)\)\(4,x^2-2xy+8x-16y=x\left(x-2y\right)+8\left(x-2y\right)=\left(x+8\right)\left(x-2y\right)\)\(5,x^2-9y^2+5x+15y=\left(x-3y\right)\left(x+3y\right)+5\left(x+3y\right)=\left(x+3y\right)\left(x-3y+5\right)\)
A = 5x2 + 5y2 + 2xy + 8x + 16y + 5
A = ( x2 + 2xy + y2 ) + ( 4x2 + 8x + 4 ) + ( 4y2 + 16y + 16 ) - 15
A = ( x + y )2 + ( 2x + 2 )2 + ( 2y + 4 )2 - 15 \(\le\)- 15
Dấu = xảy ra \(\Leftrightarrow\)2x + 2 = 0 ; 2y + 4 = 0
\(\Rightarrow\)x = - 1 và y = - 2
Max A = - 15 \(\Leftrightarrow\)x = - 1 và y = - 2
Bài 1: Chắc đề là \(a^4-4a^2+4a-1\)
Ta có: \(a^4-4a^2+4a-1=a^4-\left(4a^2-4a+1\right)=a^4-\left(2a-1\right)^2=\left(a^2-2a+1\right)\left(a^2+2a-1\right)=\left(a-1\right)^2\left(a^2+2a-1\right)\)
là số nguyên tố \(\Leftrightarrow\left[{}\begin{matrix}\left(a-1\right)^2=1\\a^2+2a-1=1\end{matrix}\right.\) ( tự giải tiếp)
Bài 2: Làm mẫu một bài thôi nhé
a) Đặt A = \(2x^2+2y^2+2xy-8x-10y+2025\)
\(2A=4x^2+4y^2+4xy-16x-20y+4050\)
\(=\left(2x\right)^2+2.2x\left(y-4\right)+\left(y-4\right)^2-\left(y-4\right)^2+4y^2-20y+4050\)
\(=\left(2x+y-4\right)^2-\left(y^2-8y+16\right)+4y^2-20y+4050\)
\(=\left(2x+y-4\right)^2+3y^2-12y+4034=\left(2x+y-4\right)^2+3\left(y^2-4y+4\right)+4022=\left(2x+y-4\right)^2+3\left(y-2\right)^2+4022\ge4022\forall x,y\)
Vậy min A = 4022 \(\Leftrightarrow\left\{{}\begin{matrix}2x+y-4=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
\(-x^2 -3y^2 -2xy +10x +16y +18 \)
\(=-x^2-2xy+10x-y^2+10y-25-2y^2+6y+43\)
\(=-\left(x^2+2xy-10x+y^2-10y+25\right)-2\left(y^2-3y-\dfrac{43}{2}\right)\)
\(=-\left[\left(x^2+2xy+y^2\right)-\left(10x+10y\right)+25\right]-2\left(y^2-3y+\dfrac{9}{4}-\dfrac{95}{4}\right)\)
\(=-\left[\left(x+y\right)^2-10\left(x+y\right)+25\right]-2\left(y^2-3y+\dfrac{9}{4}\right)+\dfrac{95}{2}\)
\(=-\left(x+y-5\right)^2-2\left(y-\dfrac{3}{2}\right)^2+\dfrac{95}{2}\le\dfrac{95}{2}\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}-\left(x+y-5\right)^2=0\\-2\left(y-\dfrac{3}{2}\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\y=\dfrac{3}{2}\end{matrix}\right.\)
\(A=x^2-4y^4=\left(x-2y^2\right)\left(x+2y^2\right)\)
\(B=8x^3+1=\left(2x+1\right)\left(4x^2-2x+1\right)\)
\(C=54x^3-16y^3=2\left(27x^3-8y^3\right)=2\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)
\(D=x^2-6x+8=\left(x^2-6x+9\right)-1=\left(x-3\right)^2-1=\left(x-3-1\right)\left(x-3+1\right)=\left(x-4\right)\left(x-2\right)\)
\(E=2x^2-5x+2=\left(2x^2-4x\right)-\left(x-2\right)=2x\left(x-2\right)-\left(x-2\right)=\left(x-2\right)\left(2x-1\right)\)
\(G=x^4+2x^2-3=\left(x^4+3x^2\right)-\left(x^2+3\right)=x^2\left(x^2+3\right)-\left(x^2+3\right)=\left(x^2+3\right)\left(x^2-1\right)=\left(x^2+3\right)\left(x-1\right)\left(x+1\right)\)